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A Thompson experiment

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Imagine that you set up an apparatus to reproduce Thomson's experiment, as shown in the figure. In an highly evacuated glass tube, a beam of electrons, each moving with speed vo, passes between two parallel plates and strikes a screen at a distance L from the end of the plates.

    First, you observe the point where the beam strikes the screen when there is no electric field between the plates. Then, you observe the point where the beam strikes the screen when a uniform electric field of magnitude is established between the plates. Call the distance between these two points delta y .

    What is the distance, delta y, between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
    Express your answer in terms of e ,m ,d ,vo ,L , and Eo .


    2. Relevant equations



    3. The attempt at a solution

    I really have no idea how to begin with it.
    I assume I need to find the distance when the beam is not deflected and then the distance after the beam is deflected. then substract them.

    Please help me..
     

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  2. jcsd
  3. Feb 3, 2008 #2

    Doc Al

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    Hint: Which way does the force act? How long does it act for? (How long is the electron in the field?)
     
  4. Feb 3, 2008 #3
    the force acts on the electron.
    F = q(VxB).
    about how long is it? does it depend on the length of the plate?
     
  5. Feb 3, 2008 #4

    Doc Al

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    It's an electric field, not a magnetic field. :wink: What direction does the force act?
     
  6. Feb 3, 2008 #5
    ups..
    F=qE
    about the force, do you mean it is perpendicular to the electric field?
     
  7. Feb 3, 2008 #6

    Doc Al

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    The force is parallel to the field, not perpendicular.
     
  8. Feb 3, 2008 #7
    can you give me another hint? I'm really clueless..
     
  9. Feb 3, 2008 #8

    Doc Al

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    Which way does the field point? (Look at the diagram!) Which way does the force act?
     
  10. Feb 3, 2008 #9
    the E points downward, so the F will point downward too..
     
  11. Feb 3, 2008 #10

    Doc Al

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    Careful. What's the sign of the electron's charge?
     
  12. Feb 3, 2008 #11
    the sign is negative charge.
     
  13. Feb 3, 2008 #12

    Doc Al

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    Right. So the force points opposite to the field. (That's why the diagram shows the electron being deflected up.)
     
  14. Feb 3, 2008 #13
    okay. the foce is going to point upward since the electron has a negative charge.
    I look up my text book, but there is not any equation for finding a distance.
     
  15. Feb 3, 2008 #14

    Doc Al

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    You have the force. Figure out the acceleration.

    Hint to find the time that the electron is in the field: Does the field affect the horizontal motion of the electron?

    You want to find the direction that the electron is moving in once it gets out of that field.
     
  16. Feb 3, 2008 #15
    F= ma
    a=F/m
    = qE/m
    v^2/r = qE/m
    V=sr of qEr/m

    then use kinematic equation = delta y = vot-1/2gt^2
    delta y = (sr of qEr/m)t-1/2gt^2

    t = (d+L)/v
    delta Y = (sr of qEr/m)*((d+L)/v)-1/2g((d+L)/v)^2

    is that correct?
     
  17. Feb 3, 2008 #16

    Doc Al

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    OK.
    Why introduce centripetal acceleration? The acceleration is upward, not radial.

    Once the electron leaves the field, what's its acceleration?

    What is the transit time for the electron through the field?
     
  18. Feb 3, 2008 #17
    I am really confused.
    to find the time is the distance / the velocity.
    I still don't get it
     
  19. Feb 3, 2008 #18

    Doc Al

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    That's right. The time = distance/velocity, for constant speed motion. You have the horizontal distance through the field and the horizontal speed. So what's the time?

    Then you can find the deflection while in the field and the final velocity as it leaves.
     
  20. Feb 3, 2008 #19
    the horizontal distance is (d+L) right?
    then the time is (d+L)/v
    how to find the deflection? is it using delta y = Vot+1/2at^2 ?
     
  21. Feb 3, 2008 #20

    Doc Al

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    That's the total horizontal distance and time. Better break this problem into two parts: the accelerated part within the plates, and the non-accelerated part after it leaves the plates.

    How long is it between the plates? What's the vertical deflection when it leaves the plates? What direction is it moving in? (You'll need the vertical speed it gains while in the field.)
     
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